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from the latter of which subtract the quintuple of the former, and there will remain

*120Ea-4b6c4d + e.

Now divide each of the equations marked thus, *, by the coefficient of its first term, and there will come out the very values of A, B, C, D, &c. above exhibited. Q. E. D.

COROLLARY.

If every term of the proposed series a, b, c, d, &c. be subtracted from the next following, the first of the remainders, a+b,—b+c,-c+d,-d+e, &c. divided by 2, gives the value of B, the coefficient of the second term of the assumed series. And, if each of the quantities thus arising be subtracted from its succeeding one, the first of the new remainders, a ·2b+c, b. ·2c+d, c 2d+e, &c. divided by 6, will be equal to C, the coefficient of the third term of the same series. In like manner, if each of these last remainders be, again, subtracted from its succeeding one, the next remainders will be,-a+3b — 3c+d, — b + 3c 3d+e, &c. whereof the first, divided by 24, gives the coefficient of the fourth term, &c. &c. Therefore, if the first remainder of the first order be denoted by P, the first of the second order by Q, the first of the third by R, the first of the fourth by S, &c. then,

P

R

S

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2 being

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Q 2.3

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2.3 4.5

=

E, &c. it is manifest that the sum of the series a + b +c+d+e+f, &c. will be truly expressed by

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Example 1. Let the sum of the series of squares 1+ 4+9+16....+n2 be required. Then, taking

the differences of the several orders, according to the preceding corollary, we have

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Example 2. Let it be required to find the sum of n terms of the following series of cubes, viz. 27 +64 + 125 +216 + 343 + 512, &c. Proceeding here, as in the last example, we have

27, 64, 125, 216, 343, 512, &c.

37, 61, 91, 127, 169, &c.

24, 30, 36, 42, &c.

6, 6, 6, &c.

0, 0, &c.

Therefore, by substituting 27 for a, 37 for P, 24 for C, and 6 for D, we thence get

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Example 3. Let the series propounded be 2+6+12+

20+30, &c.

In this case, we have

2, 6, 12, 20, 30, &c.

4, 6, 8, 10, &c.
2, 2, 2, &c.

4, Q

=

&c.

= 2, P = 2, and R, S, 0, the sum of the series will therefore be 2n + 2n xn-1 x n 2 n3 + 3n2 + 2n

Hence, a being each =

4n x n

1

+

2

6

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3

:

And in the very same manner the sum of the series may be truly found, in all cases where the differences of any order become equal among themselves and even in other cases, where the differences do not terminate, a near approximation may be obtained, by carrying on the process to a sufficient length.

SECTION XV.

Of Figurate Numbers, their Sums, and the Sums of their
Reciprocals, with other matters of the like nature.
THAT series which arises by adding together a rank
Units (called fig. No. of the 1st ord.)
Figurate numbers of the 2d order
Figurate numbers of the 3d order
Figurate numbers of the 4th order
Figurate numbers of the 5th order
Figurate numbers of the 6th order
Therefore the figurative numbers

of

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is called a series of

fig. numb. of the

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3d

4th

5th

6th

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2 3 4. 5.

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are

3

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1

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4

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6.10.15. &c.

10. 20. 35. &c.

5 15. 35.70. &c.

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7th

order.

Hence it is manifest, that, to find a general expression for a figurate number of any order, is the same thing as to find the sum of all the figurate numbers of the preceding order, so far. Let n be put to denote the distance of any such number from the beginning of its respective

order, or the number of terms in the preceding order whereof it is composed: then it is evident, by inspection, that the sum of the first order, or the nth term of the second, will be truly expressed by n, the number of terms from the beginning. It is also evident, from sect. 14, p. 203, that the sum of the second order, 1 + 2 + 3 n n+ 1

+4 n, will be

....

n2

2

n

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1

) which,

2

according to the preceding observation, is also the value of the nth term of the third order. Hence, if the numbers, 1, 2, 3, 4, 5, &c. be successively wrote instead of n,

in the general expression

9

3

2

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2 + 1, 1⁄2 + 2, 1 + 1/1, 16 + 4, 2/5 + §, &c. for the values of the first, second, third, fourth, fifth, &c. terms of this order, respectively; whence it appears, that the series 1 + 3 + 6 + 10 + 15 + 21, &c. may be resolved into these two others, viz.

1 + 1 + 2 + 1 + 25 + 36 &c. and

1 + 1/4 + 3 + 4 + § + ; &c.

The former of which being a series of squares, its sum

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and that of the latter series (by case 1, p. 203) appears

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true value of the proposed series 1 + 3 + 6 + 10 + 15, &c. continued to n terms, and therefore equal, likewise, to the nth term of the next superior order, 1 + 4 +10+20 + 35, &c. Let, therefore, 1, 2, 3, 4, 5, &c. (as above) be successively wrote for n in this ge

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neral expression, + +, and it will become

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¿ + 1 + }, } + 1⁄2 + 3, + 27 + 2 + 3, 64 + 16 +3, &c. for the values of the first, second, third, fourth, &c.

terms of the fourth order respectively; whence it appears that the series 1+4+10+20+35, &c. may be resolved into these three others, viz.

1+8+27+64 +125+216.... n3

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n4 13 n2 n3

whereof the sums are + +

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n2
+ +
24 12 24' 6 4

n

12'

(by p. 202, and 203) the aggregate of

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n+3) will consequently be the true value of the whole

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4

series. After the same manner, the sum of the fifth order

n+ 4

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X

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3

4

5

;

n n+1 n+2 n+3 will appear to be X 1 2 from whence the law of continuation is manifest. And it may not be amiss to observe here, that though the conclusions thus brought out, are derived by means of the sums of powers determined in the preceding section, yet the same values may be otherwise obtained, by a direct investigation, from either of the two general methods there laid down.

In order now to find the sum of the reciprocals of any series of figurate numbers, suppose 1+b+bc+bcd+bcde +bcdef+ &c. to be a series whose terms continually decrease, from the first to the last, so that the last may vanish, or become indefinitely small: then, by taking the excess of every term above the next following one, we shall have 1-b, b x 1-c, be x 1-d, bcd x 1-e, bcde x 1-f, &c. The sum of all which is, evidently, equal to the

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